Let us start from the classical double slit experiment.A classical double slit interference involves superposition of two coherent electric fields (secondary waves emitted from the different parts of the wavefront of the primary wave in conjunction with optical devices).Here,intensity is
$\ I_{total}\ =\ I_1\ +\ I_2\ +\ I_{12}\$
and this is equal (apart from a constant factor) to
$[\ <\vec{E}_1^2\ >_{T}\ +\ <\vec{E}_2^2\ >_{T}\ + 2\ <\vec{E}_1\cdot\vec{E}_2\ >_{T}]\$
Clearly,the first two terms are constant.The interference effect comes from the last term,namely the interference term.the interference term arises because of dotting of $\vec{E}_1\$ and $\vec{E}_2\$.
However,if the intensity of the source is so reduced that only a single photon is allowed to fall on the screen at a time,even then, the interference is observed after waiting sufficiently long time,so that our eyes can see the intensity pattern.This means,even a single photon can produce interference pattern, though our eyes may not perceive it.
This raises the question,how can a single photon produce interference...as it is believed to pass through any one of the two slits.This question was resolved in QM saying that a single photon is to be considered as linear superposition of the two states: one passing through the upper (1) pinhole,the other passing through the lower pinhole(2).A single photon is written as
$\ a\ |1>\ +\ b\ |2>\$
When there exists no evidence of which path the photon has followed,we can see the interference pattern.This interference pattern will consist of $\ <1|1>\$,$\ <2|2>\$, and $\ <1|2>\$ terms. The cross term is the interference term arising from the interaction between the two different parts of the "same photon",leaked through two holes.
Needless to say,this experiment has been successfully done with material particles (electrons) also,thus establishing the wave nature of matter.In all such cases,"the possible paths can be combined coherently (in this case the experiment does not reveal which path a photon/electron follows) resulting in an interference pattern.If the experiment reveals the path of the photon/electron,the interference pattern destroys.The meaning of "coherence" is sheerly different than in classical optics(coherence is no way an easy thing to understand,either in optics or in quantum mechanics(occurs frequently in many body QM;refer to Ballentine ch19.Later I will write something on quantum coherence when myself will be in a position to write).
Thus far,we have seen interference with the one-particle systems.These experiments have more or less direct counterpart to phenomena in classical optics.They do not show how enigmatic observations superposition principle can lead to.However,if we investigate two particle systems,because they are the next to the simplest,we have examples which have no classical counterpart and thus, the novel nature of quantum mechanics becomes clear.
A two particle quantum state will be most conveniently described by 3N=3X2=6 co-ordinates. We assign a six-dimensional configuration space to our system unlike the classical wave phenomenon.The multidimensional wavelike character of the quantum states gives coherence a richer meaning in quantum mechanics,than in classical optics.
If the two bodies in the system do not interact with one another,then,the Hamiltonian is H=H1+H2.Each body may have internal degrees of freedom;our demand is-they do not interact with that of the other body.Then,two distinct types of Schrodinger's equation comes into play: a two-body equation and two one-body equation.The consequence is that the product of one body wave functions is also a solution of the two body equation.
However, if the two particles in the system are mutually interacting,then,Hamiltonian will have an additional interaction term: H_12 and the above separation of variable method will not work.
We are now in a position to investigate the experiment described in the book:p-14.
The heart of the experiment is that "the two particle states can show interference effects if both particles are detected in co-incidence while not showing the conventional one-particle interference effects".Our experiment observes the daughter particles of the decay:
$\ A\rightarrow\ a\ +\ b\$
Our experimental set-up,we will make it possible to determine which path one particle takes by doing some measurement on the second particle;in this case neither of the two particles show interference $\ P_a\$ or $\ P_b\$,but there may be interference effect in the co-incidence rate $\ P_{ab}\$
It is important to appreciate that whether the observation is made or not does not matter;what matters is whether such an observation is possible or not.The strength of this statement will be clear shortly.
From the figure,we see the detector screens $\ D_a\$ and $\ D_b\$ are sensitive to a and b particles only.They record the co-ordinates of the particles striking them in coincidence.This is the joint probability distribution $\ P_{ab}\$.We agree to the convention that the upper pin holes would be designated (+) and the lower ones (-).
First of all,if A is at rest $\Delta\ p_A\rightarrow\ 0\$,its wave function resides across whole of the region within the two double-slit screens.In this case,the daughter particles are emitted back to back so as to conserve linear momentum.
Since we are assured that our set up enables us to determine the path followed by one state by performing some measurement on the other state,there will be no interference effect for each of them.Hence,the daughter states are to be seen as particles,not as waves.If the daughter particle behaved like waves,then, it were possible to write each of them in the form $\ c_{1}\ |+>\ +\ c_{2}\ |->\$ and interference pattern would have been obtained (provided there had existed no means to determine which path a state follows)---Our experiment does not entertain this facility.
We will make sure if $\ a\$ passes through the upper hole in the right side(+R),$\ b\$ must pass through the lower hole (-L) in the left side---i.e.$\ b\$ follows diametrically opposite path.Two jets of a and b particles rush in the opposite directions.If we name the pinholes as $\ +\ R\$,$\ -\ R\$,$\ +\ L\$ and $\ -\ L\$ then,there might be four such back to back jets of our concern: (+,+),(-,-).(+,-),(-,+).One such case(+,-) is shown below:
If the position of A were prescisely known to be at the centre of the set up,its momentum would be totally uncertain,and there will be no correlation of the directions of a and b,and the observation on any one of them cannot reveal the path taken by the other.In this case $\ I_1\$,$\ I_2\$ and $\ I_{12}\$ are all present.
All these being said,we must clarify how we do ensure that the two daughters can pass through one pair of diametrically opposite pin-holes.If particle $\ a\$ comes out from the upper hole in the right side,the corresponding particle $\ b\$ can either pass through (+L) or (-L) hole.It turns out that it depends on the size of the source.
We can look upon it this way: if the two daughter states are to pass diametrically opposite holes,i.e. A is to illuminate one or the other of the two opposed holes,then angle $\Theta\$ between their momenta (remembering $\Delta\ p_A\rightarrow\ 0\$ but $\Delta\ p_A\neq\ 0\$) must be very small compared to $\phi\$ where the latter is defined in this figure:
Note that when I drew the jets(2nd figure of the post),I drew the angle of the cone enough small.
Clearly,$\Theta\sim\frac{\Delta\ p_A}{\hbar\ k}\ll\phi\$.Here $\Delta\ p_A\$ is the uncertainty in the y component of A.This manifests as the difference angle $\Theta\$ between the momenta of the daughter particles.Again,from uncertainty of A's vertical position,$\Delta\ p_A\ge\frac{\hbar}{s}\$.From all these,we have $\ s\gg\frac{\ 1}{\phi\ k}\$
This condition being satisfied,the two daughter particles will be tracing opposite path and will reach diametrically opposite pin holes.In that case,measurement on one will definitely ensure the path traced by the other.
If this condition is not satisfied,i.e.the source size is small enough,then,$\Delta\ p_A\$ must be much bigger.This means,$\Theta\sim\frac{\Delta\ p_A}{\hbar\ k}\$ is much bigger compared to $\phi\$.In this limit,(+,+) or (-,-) decay cannot be neglected.In that case,one cannot definitely say the path traced by a particle by doing measurement on the other.
It is clear from the above figure,as the angle $\Theta\$,with the same $\Delta\ p_A\$,A may also illuminate two back to back pin-holes.Clearly, angle AOB > angle AO'B
Beyond the pin-holes,$\psi_{out}\$ is a linear combination of four terms:
$\psi_{out}\ =\ C_1\psi(\ L_a^+)\psi(\ L_b^-)\ +\ C_2\psi(\ L_a^+)\psi(\ L_b^+)\$
$\ +\ C_3\psi(\ L_a^-)\psi(\ L_b^+)\ C_4\psi(\ L_a^-)\psi(\ L_b^-)\$
Here $\psi\$ is a general spherical wave of the form $\psi\ =\frac{\ e^{\ i\ k\ r}}{r}\$
Matching the above with $\psi_{in}\$ at the boundary,we can evaluate the constants.Gottfried assumed a spherical wavefunction and the interaction to be rotation-invariant.This makes the state $\psi_A\ =\psi(\vec{r}_a\vec{r}_b)\$.From the symmetry argument in the book,we have
$\psi(\vec{r}_{\ a^+}\vec{r}_{\ b^-})\ =\psi(\vec{r}_{\ a^+}\vec{r}_{\ b^-})\ =\alpha\$ &
$\psi(\vec{r}_{\ a^+}\vec{r}_{\ b^+})\ =\psi(\vec{r}_{\ a^-}\vec{r}_{\ b^-})\ =\beta\$
And assuming that the screen is quite far off from the pin-holes,we arrive at eq (32) in the book:
$\psi_{out}\ =\alpha\ (\ e^{\ i\ k\ L_{a^+}}\ e^{\ i\ k\ L_{b^-}}\ + \ e^{\ i\ k\ L_{a^-}}\ e^{\ i\ k\ L_{b^+}}\ )\$
$\ +\beta\ (\ e^{\ i\ k\ L_{a^+}}\ e^{\ i\ k\ L_{b^+}}\ +\ e^{\ i\ k\ L_{a^-}}\ e^{\ i\ k\ L_{b^-}}\ )\$
Now,with proper substitutions,like $\ L_{\ a^+}^2\ =\ L_0^2\ +\ (\ a\ -\ y_a)^2\$ etc,we can get $\psi_{out}\$ in terms of other variables.
Since,in our experiment (+,+) and (-,-) decays are prohibited by adjusting the vertical height of the source,$\alpha\gg\beta\$ and we may safely neglect the $\beta\$ term in the expression of $\psi_{out}\$
The interference term comes out ~$\ P_{ab}\doteq\ |\ cos(\ k\theta\ [\ y_a\ -\ y_b])|^2\$
There will be a little more discussion on eqn (36) and (37) that shed more light on the experiment.I will not write right now as I am enough tired with this experiment.May be some time later I will be back at this.
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